By considering the roots of #z^5 =1#, how do you show that cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) = -1?

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Answer 1 · 2017-02-25T08:33:20

Please see below.

As #z^5=1=(cos0+isin0)#

According to De-Moivre's theorem

#z=(z^5)^(1/5)=(cos((2npi)/5)+isin((2npi)/5))# and we get five roots of #1# by changing value of #n# from #0# to #4# and we get five roots as

#cos0+isin0#or #1+i0#, #cos((2pi)/5)+sin((2pi)/5)#, #cos((4pi)/5)+sin((4pi)/5)#, #cos((6pi)/5)+sin((6pi)/5)# and #cos((8pi)/5)+sin((8pi)/5)#

and these five roots, when graphed on the complex plane, are equally spaced around a circle of unit radius as shown below.

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and it is apparent that their sum will be zero.

As such comparing real and imaginary parts we should have

#1+cos((2pi)/5)+cos((4pi)/5)+cos((6pi)/5)+cos((8pi)/5)=0#

and #0+sin((2pi)/5)+sin((4pi)/5)+sin((6pi)/5)+sin((8pi)/5)=0#

i.e. #cos((2pi)/5)+cos((4pi)/5)+cos((6pi)/5)+cos((8pi)/5)=-1#

and #sin((2pi)/5)+sin((4pi)/5)+sin((6pi)/5)+sin((8pi)/5)=0#